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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
-
How to use it?
- Canonical form:
- 2*x-3/4*x-8=0
- 4x^2+25y^2-16x+50y-67=0
- x^2+y^2-xy+x+y=0
- x^2-y^2-6x-4y+1=0
- Plot:
- (y+4)^2+(x+2)^2=25
-
(x-1)^2+(y-1)^2=4
- ((|y|)+2)^2+((|x|)-4)^2=8
- y=-x2-4*x-4
-
Identical expressions
- five x^ two +4xy+ eight y²-(sixteen / five ^(one / two))x+(8/5^(- one / two))y+ thirty-two = zero
- 5x squared plus 4xy plus 8y² minus (16 divide by 5 to the power of (1 divide by 2))x plus (8 divide by 5 to the power of ( minus 1 divide by 2))y plus 32 equally 0
- five x to the power of two plus 4xy plus eight y² minus (sixteen divide by five to the power of (one divide by two))x plus (8 divide by 5 to the power of ( minus one divide by two))y plus thirty minus two equally zero
- 5x2+4xy+8y²-(16/5(1/2))x+(8/5(-1/2))y+32=0
- 5x2+4xy+8y²-16/51/2x+8/5-1/2y+32=0
- 5x²+4xy+8y²-(16/5^(1/2))x+(8/5^(-1/2))y+32=0
- 5x to the power of 2+4xy+8y²-(16/5 to the power of (1/2))x+(8/5 to the power of (-1/2))y+32=0
- 5x^2+4xy+8y²-16/5^1/2x+8/5^-1/2y+32=0
- 5x^2+4xy+8y²-(16/5^(1/2))x+(8/5^(-1/2))y+32=O
- 5x^2+4xy+8y²-(16 divide by 5^(1 divide by 2))x+(8 divide by 5^(-1 divide by 2))y+32=0
-
Similar expressions
- 5x^2+4xy+8y²-(16/5^(1/2))x-(8/5^(-1/2))y+32=0
- 5x^2-4xy+8y²-(16/5^(1/2))x+(8/5^(-1/2))y+32=0
- 5x^2+4xy+8y²-(16/5^(1/2))x+(8/5^(1/2))y+32=0
- 5x^2+4xy+8y²+(16/5^(1/2))x+(8/5^(-1/2))y+32=0
- 5x^2+4xy-8y²-(16/5^(1/2))x+(8/5^(-1/2))y+32=0
- 5x^2+4xy+8y²-(16/5^(1/2))x+(8/5^(-1/2))y-32=0
5x^2+4xy+8y²-(16/5^(1/2))x+(8/5^(-1/2))y+32=0 canonical form
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The solution
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___ ____
2 2 4*x*\/ 5 y*\/ 10
32 + 5*x + 8*y + 4*x*y - --------- + -------- = 0
5 4
$$5 x^{2} + 4 x y + 8 y^{2} - \frac{4 \sqrt{5} x}{5} + \frac{\sqrt{10} y}{4} + 32 = 0$$
5*x^2 + 4*x*y - 4*sqrt(5)*x/5 + 8*y^2 + sqrt(10)*y/4 + 32 = 0
Detail solution
Given line equation of 2-order:
$$5 x^{2} + 4 x y + 8 y^{2} - \frac{4 \sqrt{5} x}{5} + \frac{\sqrt{10} y}{4} + 32 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = - \frac{2 \sqrt{5}}{5}$$
$$a_{22} = 8$$
$$a_{23} = \frac{\sqrt{10}}{8}$$
$$a_{33} = 32$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 2\\2 & 8\end{matrix}\right|$$
$$\Delta = 36$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$5 x_{0} + 2 y_{0} - \frac{2 \sqrt{5}}{5} = 0$$
$$2 x_{0} + 8 y_{0} + \frac{\sqrt{10}}{8} = 0$$
then
$$x_{0} = \frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}$$
$$y_{0} = - \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - \frac{2 \sqrt{5} x_{0}}{5} + \frac{\sqrt{10} y_{0}}{8} + 32$$
$$a'_{33} = - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32$$
then The equation is transformed to
$$5 x'^{2} + 4 x' y' + 8 y'^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$5 x'^{2} + 4 x' y' + 8 y'^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
to
$$8 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 5 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
simplify
$$4 \tilde x^{2} + 9 \tilde y^{2} - \frac{\sqrt{2}}{36} + \frac{61057}{1920} = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{3}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_{2} = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
$$5 x^{2} + 4 x y + 8 y^{2} - \frac{4 \sqrt{5} x}{5} + \frac{\sqrt{10} y}{4} + 32 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = - \frac{2 \sqrt{5}}{5}$$
$$a_{22} = 8$$
$$a_{23} = \frac{\sqrt{10}}{8}$$
$$a_{33} = 32$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 2\\2 & 8\end{matrix}\right|$$
$$\Delta = 36$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$5 x_{0} + 2 y_{0} - \frac{2 \sqrt{5}}{5} = 0$$
$$2 x_{0} + 8 y_{0} + \frac{\sqrt{10}}{8} = 0$$
then
$$x_{0} = \frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}$$
$$y_{0} = - \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}$$
Thus, we have the equation in the coordinate system O'x'y'
$$a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} + a'_{33} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - \frac{2 \sqrt{5} x_{0}}{5} + \frac{\sqrt{10} y_{0}}{8} + 32$$
$$a'_{33} = - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32$$
then The equation is transformed to
$$5 x'^{2} + 4 x' y' + 8 y'^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{- \cos^{2}{\left(\phi \right)} + 1}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
substitute coefficients
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
then the equation turns from
$$5 x'^{2} + 4 x' y' + 8 y'^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
to
$$8 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 5 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - \frac{2 \sqrt{5} \left(\frac{\sqrt{10}}{144} + \frac{4 \sqrt{5}}{45}\right)}{5} + \frac{\sqrt{10} \left(- \frac{5 \sqrt{10}}{288} - \frac{\sqrt{5}}{45}\right)}{8} + 32 = 0$$
simplify
$$4 \tilde x^{2} + 9 \tilde y^{2} - \frac{\sqrt{2}}{36} + \frac{61057}{1920} = 0$$
Given equation is imaginary ellipse
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{3}\right)^{2}} = -1$$
- reduced to canonical form
The center of canonical coordinate system at point O
____ ___ ____ ___ \/ 10 4*\/ 5 5*\/ 10 \/ 5 (------ + -------, - -------- - -----) 144 45 288 45
Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_{2} = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$
Invariants method
Given line equation of 2-order:
$$5 x^{2} + 4 x y + 8 y^{2} - \frac{4 \sqrt{5} x}{5} + \frac{\sqrt{10} y}{4} + 32 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = - \frac{2 \sqrt{5}}{5}$$
$$a_{22} = 8$$
$$a_{23} = \frac{\sqrt{10}}{8}$$
$$a_{33} = 32$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 13$$
$$I_{3} = \left|\begin{matrix}5 & 2 & - \frac{2 \sqrt{5}}{5}\\2 & 8 & \frac{\sqrt{10}}{8}\\- \frac{2 \sqrt{5}}{5} & \frac{\sqrt{10}}{8} & 32\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 5 & 2\\2 & - \lambda + 8\end{matrix}\right|$$
$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = - \sqrt{2} + \frac{183171}{160}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = \frac{66407}{160}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 13 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$9 \tilde x^{2} + 4 \tilde y^{2} - \frac{\sqrt{2}}{36} + \frac{61057}{1920} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{3}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{2}\right)^{2}} = -1$$
- reduced to canonical form
$$5 x^{2} + 4 x y + 8 y^{2} - \frac{4 \sqrt{5} x}{5} + \frac{\sqrt{10} y}{4} + 32 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = - \frac{2 \sqrt{5}}{5}$$
$$a_{22} = 8$$
$$a_{23} = \frac{\sqrt{10}}{8}$$
$$a_{33} = 32$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|substitute coefficients
$$I_{1} = 13$$
|5 2|
I2 = | |
|2 8|$$I_{3} = \left|\begin{matrix}5 & 2 & - \frac{2 \sqrt{5}}{5}\\2 & 8 & \frac{\sqrt{10}}{8}\\- \frac{2 \sqrt{5}}{5} & \frac{\sqrt{10}}{8} & 32\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 5 & 2\\2 & - \lambda + 8\end{matrix}\right|$$
| ___| | ____|
| -2*\/ 5 | | \/ 10 |
| 5 --------| | 8 ------|
| 5 | | 8 |
K2 = | | + | |
| ___ | | ____ |
|-2*\/ 5 | |\/ 10 |
|-------- 32 | |------ 32 |
| 5 | | 8 |$$I_{1} = 13$$
$$I_{2} = 36$$
$$I_{3} = - \sqrt{2} + \frac{183171}{160}$$
$$I{\left(\lambda \right)} = \lambda^{2} - 13 \lambda + 36$$
$$K_{2} = \frac{66407}{160}$$
Because
$$I_{2} > 0 \wedge I_{1} I_{3} > 0$$
then by line type:
this equation is of type : imaginary ellipse
Make the characteristic equation for the line:
$$- I_{1} \lambda + \lambda^{2} + I_{2} = 0$$
or
$$\lambda^{2} - 13 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
then the canonical form of the equation will be
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
or
$$9 \tilde x^{2} + 4 \tilde y^{2} - \frac{\sqrt{2}}{36} + \frac{61057}{1920} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{3}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{- \frac{\sqrt{2}}{36} + \frac{61057}{1920}}}{2}\right)^{2}} = -1$$
- reduced to canonical form
The graph